Logarithm of Power/Natural Logarithm/Proof 2

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Theorem

Let $x \in \R$ be a strictly positive real number.

Let $a \in \R$ be a real number such that $a > 1$.

Let $r \in \R$ be any real number.

Let $\ln x$ be the natural logarithm of $x$.


Then:

$\map \ln {x^r} = r \ln x$


Proof

\(\ds \ln a\) \(=\) \(\ds b\)
\(\ds \leadstoandfrom \ \ \) \(\ds e^b\) \(=\) \(\ds a\)
\(\ds \leadsto \ \ \) \(\ds \paren {e^b}^c\) \(=\) \(\ds a^c\)
\(\ds \leadsto \ \ \) \(\ds e^{c b}\) \(=\) \(\ds a^c\) Exponential of Product
\(\ds \leadsto \ \ \) \(\ds \ln e^{c b}\) \(=\) \(\ds \ln a^c\)
\(\ds \leadsto \ \ \) \(\ds c b\) \(=\) \(\ds \ln a^c\) Exponential of Natural Logarithm


By hypothesis, $\ln a = b$.

Multiplying both sides by $c$:

$c \ln a = c b$

But we proved above that:

$c b = \ln a^c$

$\blacksquare$


Sources